¿Hay una distribución conjunta paramétrica tal que$X$ y$Y$ son uniformes y$\mathbb{E}[Y \;|\; X]$ es lineal?

Question

Hay una paramétrico de la distribución conjunta tal que $X$ $Y$ son tanto uniforme en $[0, 1]$ (es decir, una cópula) y $\mathbb{E}[Y | X = x]$ es lineal (es decir afín) en $x$? Es decir, $$\mathbb{E}[Y \;|\; X = x] = a + b\,x$$ mientras que $X$ $Y$ son cada marginalmente $\text{Uniform}[0, 1]$.

Por supuesto, yo podría dejar el $X$ $Y$ ser independientes, en cuyo caso $\mathbb{E}[Y \;|\; X = x] = 0.5$. ¿Hay alguna simple paramétrico cúpulas que satisfacer mis limitaciones, sin que las dos variables sean independientes? (Otro caso extremo sería $Y=X$$\mathbb{E}[Y \;|\; X = x] = x$.)

Tenga en cuenta que el$a$$b$$\mathbb{E}[Y \;|\; X = x] = a + b\,x$, tiene un grado de libertad, ya que $\mathbb{E}[Y] = 0.5 = \mathbb{E}[\mathbb{E}[Y \; | \; X]] = a + 0.5\,b$.

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Algunos en el mundo real motivación para hacer esto menos abstracto: Chetty et al investigación sobre intergeneracional de la movilidad de la renta se encuentra (aproximadamente) lineal ingresos cúpulas (es decir, aproximadamente lineal de la relación entre los ingresos de los padres rango y su ingreso del niño rango). Ver http://www.rajchetty.com/chettyfiles/mobility_trends_published.pdf (Figura 1. Los Ingresos de los menores de Rango frente a los Ingresos de los Padres de Rango por Cohorte de Nacimiento) y http://www.rajchetty.com/chettyfiles/mobility_geo.pdf (Figura II: la Asociación entre los Niños y los Padres' Rangos Percentiles).

Answer 1

Podemos desarrollar rica paramétrico de las familias de la solución trivial con cópula $F(x,y) = \min(x,y)$, en el caso de perfecto (positivo) de correlación, y su contraparte para una perfecta correlación negativa. Concentrar la probabilidad de lugar a lo largo del segmento de línea que conecta $(0,\alpha)$ $(1,\beta)$ $\beta\gt \alpha$da la cópula

$$F(x,y;\alpha,\beta) = \casos{\matriz{x, y, y 0\le s \lt \alpha\text{ o }\beta \lt y \le 1 \\ \beta x, y x(\beta\alpha)\le y-\alpha \\ \alpha x + y-\alpha&\text{lo contrario.}}}$$

A similar copula arises when $\beta \lt \alpha$, which I will also designate $F(x,y;\alpha,\beta)$.

Think of these as mixtures: when $\beta \gt \alpha$, there are uniform components on the horizontal rectangles $[0,1]\times [0,\alpha]$, $[0,1]\veces[\beta,1]$, and on the central rectangle $[0,1]\veces[\alpha,\beta]$ there is a perfect correlation (whose distribution is that of $(U, \alpha+(\beta-\alpha)U)$ for a uniformly distributed variable $U$). This conception of $F$ makes it easy to compute the regression: it's a weighted sum of the three conditional means,

$$\mathbb{E}(Y\mid X) = \alpha\left(\frac{\alpha}{2}\right) + (\beta-\alpha)\left(\alpha + (\beta-\alpha)X\right) + (1-\beta)\left(\frac{1+\beta}{2}\right).$$

This evidently is linear in $X$: the intercept equals $(1+(\beta-\alpha)^2)/2$ and the slope is $(\beta-\alpha)^2$ times the sign of $\beta-\alpha$. Moreover, it has been constructed to have uniform marginals.

To create a parametric family, choose any parametric distribution for $(\alpha,\beta)$ with parameter $\theta$. Let $G(\alpha,\beta;\theta)$ be the distribution function. It describes a mixture of the $F(;\alpha,\beta)$ via integration:

$$\tilde F(x,y;\theta) = \iint F(x,y;\alpha,\beta)dG(\alpha,\beta;\theta)$$

is the distribution function (copula). Because each $F(;\alpha,\beta)$ has uniform marginals, so does $\tilde F(;\theta)$. Moreover, its regression is linear because

$$\eqalign{ \mathbb{E}_{\tilde F(;\theta)}(Y\mid X) &= \iint \mathbb{E}_{F(;\alpha,\beta)}(Y\mid X)dG(\alpha,\beta;\theta)\\ &=\iint ((1+(\beta\alpha)^2)/2 + \operatorname{sgn}(\beta\alpha)(\beta\alpha)^2 X)dG(\alpha,\beta;\theta) \\ &= \iint (1+(\beta\alpha)^2)/2 dG(\alpha,\beta;\theta) + \iint \operatorname{sgn}(\beta\alpha)(\beta\alpha)^2 dG(\alpha,\beta;\theta)\,X\\ &= \mathbb{E}_{G(;\theta)}((1+(\beta\alpha)^2)/2) + \mathbb{E}_{G(;\theta)}(\operatorname{sgn}(\beta\alpha)(\beta\alpha)^2)X. }$$

This shows how the intercept and slope are the expectations of the intercept and slope (with respect to $G$), providing useful information for selecting appropriate families $G(;\theta)$.

These graphics document a simulation from one such family. Here, $\alpha$ was drawn from a Beta$(5,1)$ distribution and $\beta$ was drawn independently from a Beta$(3,10)$ distribution. The first column shows histograms of the realizations of these parameters. The second column shows histograms of the marginal distributions of $X$ and $Y$: they are satisfactorily close to uniform. The rightmost column shows a random subset of the 100,000 simulated values, along with an estimate of its regression (the red line) and an approximation to the theoretical regression (black dotted line): they agree closely. The estimated regression was obtained by computing the means of $X$ and $Y$ within windows of $X$, then smoothing their trace with Loess.

(The "theoretical" regression line is only an approximation obtained by replacing $\alpha$ and $\beta$ in the expectation formulas by their expectations. Exact formulas are straightforward to work out in this case, but are long and messy to code.)

The R code that produced this figure can readily be used to study other families $G(;\theta)$.

#
# Draw `n` variates from the mixture copula.
# `alpha` and `beta` are intended to be realizations of G(;theta).
#
runif.xy <- function(n, alpha=0, beta=1) {
  a <- pmin(alpha, beta)
  b <- pmax(alpha, beta)
  xy <- matrix(runif(2*n), nrow=2)              # Start with a uniform distribution
  i <- xy[2,] > a & xy[2,] < b                  # Select the middle rectangle
  xy[2, i] <- (xy[1,]*(beta - alpha) + alpha)[i]# Create perfect correlation
  return(xy)
}
#
# Specify the parameters ("theta").
#
a.alpha <- 5
b.alpha <- 1
a.beta <- 3
b.beta <- 10
#
# Draw the slope `beta` and intercept `alpha` from G(;theta).
#
n.sim <- 1e5
alpha <- rbeta(n.sim, a.alpha, b.alpha)
beta <- rbeta(n.sim, a.beta, b.beta)
#
# Draw (X,Y) from the mixture.
#
sim <- runif.xy(n.sim, alpha, beta)
#
# Plot histograms of alpha, beta, X, Y.
#
par(mfcol=c(2,3))
hist(alpha); abline(v=a.alpha/(a.alpha+b.alpha), col="Red", lwd=2)
hist(beta); abline(v=a.beta/(a.beta+b.beta), col="Red", lwd=2)
hist(sim[1,], main="X Marginal", xlab="X")
hist(sim[2,], main="Y Marginal", xlab="Y")
#
# Plot the simulation and its regression curve.
#
i <- sample.int(n.sim, min(5e3, n.sim)) # Limit how many points are shown
plot(t(sim[, i]), asp=1, pch=19, col="#00000002", main="Simulation",
     xlab="X", ylab="Y")

library(zoo)
i <- order(sim[1,])
x <- as.vector(rollapply(ts(sim[1, i]), ceiling(n.sim/100), mean))
y <- as.vector(rollapply(ts(sim[2, i]), ceiling(n.sim/100), mean))
lines(lowess(y ~ x), col="Red", lwd=2)
#
# Overplot the theoretical regression curve.
#
a <- a.alpha / (a.alpha + b.alpha) # Expectation of `alpha`
b <- a.beta / (a.beta + b.beta)    # Expectation of `beta`
intercept <- (1 + (b-a)^2)/2
slope <- (b - a)^2 * sign(b-a)
abline(c(intercept, slope), lty=3, lwd=3)